@user21820 hello! Hope you’re doing good.

how can we expand floor (kx)? Where k is any integer. Say for example if we want to compute $$\int_{2}^{5}floor~(7 x)~dx $$ how can we do it?

@Knight You'd have to do the usual. floor(x) is the integer part, meaning that you decompose x = m+y where m is an integer and 0 ≤ y < 1. So you'd have to do the same here.

kx = m+y where ...

That means x = ...

So you need to break the interval of the integral into the pieces corresponding to how it breaks as deduced above.

For example for floor(x), it is usually constant and jumps only when x reaches an integer.

Similarly for floor(kx), it is usually constant and jumps only when ...?

@user21820 kx reaches an integer.

@Knight Yea so it means when x reaches ...?

That would tell you how to split the interval of the integral.

@user21820 integer/k

Right!

So do you know how to break the integral accordingly now?

Let me try

$$\int_{2}^{3} floor(7x) dx + \int_{3}^{4} floor~(7x) dx +\int_{4}^{5} floor~(7x) dx$$

(I don’t think I’m right)

$$ x = m +y \\ 7x = 7m + 7y $$ now writing $7x = n + y_1$ and $n$ can vary from 0 to 7, but I don’t know the exact intervals.

Hmmm... I think I got it.

$$ \int_{2}^{2.2} 14 dx + \int_{2.2}^{2.3}15 dx + \int_{2.3}^{2.4} 16 dx +\cdots $$

@user21820

@Knight: please try to divide interval as explained by @user21820b:

$7x$ is an integer for $x=2$. Next values of $x$ which makes $7x$ integer are given by $7x=15,16,\dots 21$ so divide interval accordingly.

@ParamanandSingh: Thanks for chiming in! I'm busy doing something else so it's great if you can continue. =)

@user21820: no issues! Please carry on with your work. I am waiting for @Knight response.

Do I have to consider all the integers between 14 and 35?

@ParamanandSingh

Yes you consider all of them @Knight

@ParamanandSingh So, for floor (7x) = 15 we have to have x = 15/7, and similar we keep on going like that?

Yes now you got it!!! @Knight

But don't do the calculations in a difficult manner. Just note that each of the split intervals is of length $1/7$ and function value is like 14,15,...,34

Thank you so much @ParamanandSingh and @user21820.

So area is $(1/7)(14+15+\dots +34)$

Wow!

@Knight I hope you know how to sum arithmetic progression

Yes.

@Knight: Great work!

Thanks

Problem: Let $$f(x) =\begin{cases} [x] & -2 \leq x \leq -1/2 \\ 2x^2 -1 & -1/2 \leq x \leq 2 \\ \end{cases} $$ [x] denotes the floor of x.

Find the number of points where $f|x| $ is discontinuous.

@user21820 I don’t know but I think $f|x| $ is same as values of $f$ at all the positive $x$.

And hence $f|x| = f(x) = 2x^2 -1$ (0 < x <2)

(the exam paper is writing $f|x|$ instead of $f(|x|)$, so let’s agree with what examine meant)